LECTURE 3: INTERPHASE MASS TRANSFER AND DIFFUSION COEFFICIENT
Dr Aradhana Srivastava, Associate Professor and Group Leader, Chemical engineering Group, BITS Pilani, Hyderabad Campus
INTERPHASE MASS TRANSFER So far we only discussed mass transfer of species in single phase (gas or liquid) Many mass-transfer operations involve the contact of two insoluble phases to permit mass transfer In such cases, the thermodynamic equilibrium between phases is important Mass transfer is derived by the deviation from the equilibrium state
If equilibrium between the phases is established, there would be no net mass transfer
2
EQUILIBRIUM RELATION There are equations that describes the equilibrium relation between the concentration of certain component in the liquid and gas phase at certain temperature and pressure Raoult’s Law(gas-liq mixture)
p A
A x A P A
total pressure pressure
vapor pressure pressure
p A
y A
P
Henry’s Law (gas-liq dilute solution) p A
y A P
partial partial pressure pressure
y A P
x A
A
P A
activity coefficient
Hx A
Distribution-law (liquid-liquid) c A, liquid 1
Kc A, liquid 2 3
DIFFUSION BETWEEN PHASES Consider the absorption of ammonia from a mixture with air using liquid water, water, in a wetted wall column The ammonia-air mixture enters the column from the bottom and flow upward and water flow downward on the inner wall of the column
Water
The concentration of the ammonia in the mixture decreases as it flow upward the concentration of the ammonia in water increases as it flow downward NH3-H2O under steady state conditions, the concentration at NH3-Air any point of the column does not change with time 4
TWO-RESISTANCE THEORY
The ammonia (solute) diffuses from the gas phase to the liquid through an interface
There is a concentration gradient in the direction of mass transfer in each phase On concentrations on the interface (yA,i and xA,i, are assumed to be in yA, G equilibrium) This simply means the mass transfer resistance is only in the fluid phase and no resistance across the interface
xA,i I
n yA,i t e
Bulk NH3-Air
r f a c e
xA, Water 5
PHASE TRANSFER COEFFICIENTS
yA, G
NNH3 can be expressed in terms of k-type coefficients: N A
k y y A ,G
y A ,i
k x x A ,i
k x
y A ,G
y A ,i
k y
x A ,L
x A ,i
xA,i I
n yA,i t e
x A ,L
Wher e :
Bulk NH3-Air
xA,
r f a c e
Water
yA,G
y A ,i is the interface concentration in the gas phase side x A ,i is the interface concentration in the liquid phase side k y is t he gas k x is the liquid
p hase m ass tr ans fer coe fficien t phase mass transfer coefficient
yA,i
yA,i= f(xA,i) xA,L
xA,i
6
OVERALL-MASS TRANSFER COEFFICIENT The interface concentration is hard to accurately measure The flux can be estimated in terms of overall mass transfer coefficient as follow
N A
K
y
y A ,G
y
* A
K
x
x
* A
x A ,L
yA, G xA,i
I n
yA,i t e
Bulk
r f a c NH3-Air e
xA,G Water
yA,G
W he re : * y A is in eq uilib rium w ith x A ,L
yA,i
* x A is in equilibriumwith y A ,G
yA* xA,L
xA,i
xA*
7
OVERALL AND INDIVIDUAL PHASE COEFFICIENT
If the equilibrium relation is linear
y A ,i
m x A ,i
Dilute solution where Henry’s Law applies * y A
N A 1 K y
m x A ,L ; K
yA *
y A ,G
m x *A
*
y A ,G
y
y A ,G
yA
y A ,G
NA
y A ,i
NA
1
1
m
K y
ky
kx
y A ,i
*
yA
NA
Similarly
y A ,G
m x A ,i
y A ,i
NA
x A ,L
N A
1
1
1
K x
mk y
kx
1
1
Resistance in gas phase
k y
Resistance in liquid phase
Total Resistance in both phases
1
Total Resistance in both phases
K y
k x 1 K X 8
EFFECT OF THE GAS SOLUBILITY IN THE MASS TRANSFER COEFFICIENTS
For highly soluble gas (the slope of the equilibrium line, m, is small)
The major resistance is in the gas phase 1
1
m
1
K k k k For low solubility gas (the slope of the equilibrium line, m, is large) y
y
x
y
yA
The major resistance is in the liquid phase XA 9
1
1
1
1
K x
mk y
kx
kx
EXAMPLE In an experimental study of the absorption of NH 3 by water in a wetted-wall column, the value of K G was found to be 2.75×10-6 kmol/(m2-s-kPa). At one point in the column, the composition of the gas and liquid phases were 8.0 and 0.115 mol% NH3, respectively. The temperature was 300 K and the total pressure was 1 atm. 85% of the total resistance to mass transfer was found to be in the gas phase. At 300° K, NH3-water solution follows Henry’s law up to 5 mol% NH3 in the liquid with m = 1.65 when the total pressure is 1 atm. Calculate: 1. Flux
of NH3 2. Individual film coefficients 3. Interfacial Compositions (y A,i and xA,i) 10
SOLUTION
Given
.
T=300 K; P=1 atm KG= 2.75x10-6 kmol/m2-s-kPa yA,G=0.080; xA,L=0.00115 . 1 0.85
N A
k y
K y
K
y
y A ,G
y *A
K y =K G P = 2.75x10 -6 x 101.3 = 2.786x10-4 kmol/m 2 -s * y A
mx A ,L
N A
K y y A ,G
1.65 * 0.00115 y *A
2.786 x10
4
1.886 x10 0.08
3
0.001886
2.18 x10
5
kmol / m 2 s 11
SOLUTION 1
0.85
y
K y
.
k
K y
k y
. K1 m . k
y
x
0.85
.
4
3.28 x10
0.85
1
m
ky 1
kx 1
1
0.85
0.15
Ky
ky
Ky
Ky
K y
k x
N A
2.786 x10
mK y 0.15
k y y A ,G y A ,i
x A ,i
1.64x 2.75 x10
y A ,G
0.15
4
kmol / m 2 .s
6
3.05 x10
3
2
kmol / m .s
y A ,i N A k y
0.08
y A ,i
0.01362
m
1.64
2.18 x10
5
3.28 x10
4
8. 305 x10
0.01362
3
12
PROBLEM The solubility of gaseous substance (Mol Wt. 26) in water is given by Henry’s law: pA = 105 xA, pA in mm of Hg. Convert the equilibrium relation to the following forms: (a) yA = m xA if the total pressure is 10 bar; (b) p A =m’CA, where CA is in gmol/litre. Also write down the equilibrium relation using the mole ratio unit. Assume that solution is dilute and has a density equal to that of water (= 1000 kg/m3)
SOLUTION
PROBLEM In a certain equipment used for absorption of SO2 from air by water at one section, the gas and liquid phase concentration of the solute are 10 mole % and 4 mass % respectively. The solution density is 6.1lb/ft3. Ata given temperature (40 0C) and pressure (10 atm), the distribution of SO2 beetween air and water can be approximately described as pA = 25 xA, where is the partial pressure of SO 2 in the gas phase in atm. The individual mass transfer coefficient are k x= 10 kmol/hm2(delta x) and k y = 8 kmol/hm2(delta y). Calculate the overall coefficient K G in kmol/hm2(delta p in mm Hg) and x Ai and yAi at the gas-liquid interface.
SOLUTION
LOCAL MASS TRANSFER COEFFICIENTS
For general case
Diffusion of more than one species No equimolar counter diffusion The mass transfer rates are large
k-type diffusion coefficients cannot be used F-type diffusion coefficient has to be used. General approach is same for finding expression
N A
A ,G
FG ln
A ,G A ,G
y A ,i A ,L
y A ,G
FL ln
A ,L
x A ,L
A ,L
x A ,i
A ,L FL
A ,G
y A ,i
A ,L
x A ,L
A ,G
y A ,G
A ,L
x A ,i
A ,G FG
Psi values are molal ratio of a component to the total moles, and F values are the mass transfer coefficients
17
EXAMPLE A wetted-wall absorption tower is fed with water as the wall liquid and an ammonia air mixture as the central-core gas. At a particular point in the lower, The ammonia concentration in the bulk gas is 0.60 mole fraction, that in the bulk liquid is 0.12 mole fraction. The temperature is 300 K and the pressure is 1 atm. Ignoring the vaporization of water, calculate the local ammonia mass-transfer flux. The rates of flow are such that F L = 0.0035 kmol/m2-s. and FG = 0.0020 kmol/m2-s. The equilibrium-distribution data for the system at 300° K and 1 atm is: y A
10.51
A
xA ;
A
0.156
0.622 x A 5.765 x A
1 ;
xA
0.3 18
SOLUTION
Given
T=300 K; P=1 atm
yA,G=0.60; xA,L=0.12 FL = 0.0035 kmol/m2-s; . FG = 0.0020 kmol/m2-s
Although this is a diffusion of A through stagnant B, the ammonia concentration is too high to use k-type mass transfer coefficient, F-type coefficient must be used y A
10.51
A
A ,G
A ,L
N A
N A
FG ln
xA ;
0.156
A
y A ,i
1
y A ,G
1 ;
xA
0.3
1
A ,G FG ln
1
0.622 x A 5.765 x A
A ,G
y A ,i
A ,G
y A ,G
FL ln
1
x A ,L
1
x A ,i
A ,L FL ln
A ,L
x A ,L
A ,L
x A ,i F L
y A ,i
1
1
y A ,G
1
x A ,L
1
x A ,i
FG
19
SOLUTION F L
y A ,i
1
1
y A ,G
1
x A ,L
1
x A ,i
y A
10.51
y A
10.51 x A 0.156
A
xA ;
A
FG
y A ,i
0.156
0.6
1.75
0.88 1
x A ,i
0.622 x A 5.765 x A
0.622 x A 5.765 x A
1
1 ;
;
xA xA
0.3
0.3
0.9
Graphically or numerically
0.8 0.7
yA,i=0.49; xAi=0.23
0.6 0.5 A
y
N A
FG ln 2 x 10
1 1 3
0.4
y A ,i
0.3
y A ,G ln
=0.4.7 x 10
4
0.2
1
0.49
0.1
1
0.6
0
km ol / m 2 s
0
0.05
0.1
0.15
0.2
0.25
0.3
xA
20
MATERIAL BALANCE
Consider SS mass transfer operation involves countercurrent contact of two immiscible phases
V is total moles of phase V Vs is moles of A-free V L is total moles of phase L Ls is moles of A-free L y is mole fraction of component A in V x is mole fraction of component A in L Y is the moles of A per mole A-free V X is mole of component A per mole Afree L Y
y 1
y
X
A = solute L2, x2, X2
V2, y2, Y2
Z=z2
L, x, X
V, y, Y
Z=z1
L1, x1, X1
V1, y1, Y1
x 1
x
21
MATERIAL BALANCE: COUNTER FLOW
Mole balance around the column moles of A entering
moles of A leaving
the column
the column
V1 y 1
LS X 2
V SY 2
Lx
Vy
L1 x 1
L, x, X
LS X
V SY
z V, y, Y
LS X 1 Z=z1
L1 x 1
Using solute free basis V SY 1
V2, y2, Y2
Z=z2
Mole balance around plane z V1 y 1
V2 y 2
Using solute free basis V SY 1
L2 x 2
L2, x2, X2
LS X 1
L1, x1, X1
V1, y1, Y1
22
MATERIAL BALANCE: COUNTER FLOW
L2, x2, X2
The mole balance in terms of solute free basis can be expressed as: L, x, X
L S
Y1
Y 2
VS
X1
X 2
or
L S
Y1
Y
VS
X1
X
Z=z2
z V, y, Y
Z=z1 L1, x1, X1
Transfer from phase V to phase L (Absorption)
V2, y2, Y2
V1, y1, Y1
Transfer from phase L to phase V (Stripping)
equilibrium curve equilibrium curve
Y1
Y
operating Line Y2
Ls/Vs
Y operating Line
Y2
Slope=Ls/Vs
Y1
23
X2
X1
X1
X2
MATERIAL BALANCE: COCURRENT FLOW
Mole balance around the column moles of A entering
moles of A leaving
the column
the column
V1 y 1
Vy
Lx
V2 y 2
Ls X 1
V SY
LS X
V SY 2
Z=z2
L, x, X
Y2
L S
Y1
Y
X 1
X2
VS
X1
X
z V, y, Y
LS X 2
Using the operating line approach: Y1
V2, y2, Y2
L2x2
Using solute free basis: V SY 1
L1 x 1
L2, x2, X2
Z=z1
L1, x1, X1
V1, y1, Y1
24
MATERIAL BALANCE: COCURRENT FLOW L2, x2, X2
The mole balance in terms of solute free basis can be expressed as: Y1
Y2
L S
Y1
Y
X 1
X2
VS
X1
X
Transfer from phase V to phase L (Absorption)
Z=z2
L, x, X
z V, y, Y
Z=z1 L1, x1, X1
V1, y1, Y1
Transfer from phase L to phase V (Stripping)
equilibrium curve
Y1
V2, y2, Y2
equilibrium curve
operating Line
Slope=-Ls/Vs
Y
Y
operating Line
Y2 Y2
Slope=-Ls/Vs Y1
X1
X
25
MATERIAL BALANCE
Masses, mass fraction, and mass ratio can be substituted consistently for moles, mole fractions, and mole ratios in the mass balance equations
Counter flow
L'S
Y 1'
Y 2'
V S'
X 1'
X '2
Cocurrent Flow '
'
Y 2
'
X 2
LS
Y1
'
X1
VS
'
'
Where the prime (´) indicates mass based property 26
EXAMPLE: ADSORPTION OF NO2 ON SILICA GEL NO2 produced by thermal process for fixation of nitrogen, is to be removed from dilute mixture with air by adsorption on silica gel in a continuous adsorber. The mass flow rate of the gas entering the adsorber is 0.5 kg/s. It contains 1.5% NO 2, by volume, and 85% of NO 2 is to be removed. Operation to be isothermal at 298° K and 1 atm. The entering gel will be free of NO 2. the equilibrium adsorption data at this temperature are: pNO2, mmHg Solid conc. (m), kg NO2/ 100 kg gel
0
2
4
6
8
10
12
0
0.4
0.9
1.65
2.60
3.65
4.85
If twice the minimum gel rate is to be used, calculate the gel mass flow rate and the composition of the gel leaving the process for: i. Counter flow operation ii. Cocurrent flow operation 27
SOLUTION: Countercurrent
1.
Plot the equilibrium data
Since the equilibrium data is given in terms of mass ratios, it is easier to use mass based equation in this case The partial pressure data have to be converted to mass ratio data (Y’) Y i Y i'
Gel
y i 1 Yi
yi
pi P
Gel/NO2
Cocurrent Gel+NO2
Air/NO2
Air/NO2
Gel
p i
M NO 2 M A ir
Air/NO2
Air/NO2
p i 760
pi
x
46
kg NO 2
29
kg A ir
Now the equilibrium data become: Xi’, kg NO2/ 100 kg gel Yi’, kgNO2/ 100 kg Air
0
0.4
0.9
1.65 2.60 3.65 4.85 28
CONTD. Y 1'
Y1
Y 2' X
' 2
M NO 2 M Air
y 1 1
46 x y 1 29
’ ' 1(1-0.85) 0.15*Y1xY 1
0.00
0.015 1
0.015
*
46 29
0.0242
0.15 * 2.42
kg NO2
*100
kg Air
kg NO 2 0.0036 kg Air
0.03
kg NO 2
'
Y1
equilibrium curve
0.025
kg gel 0.02
Minimum gel rate
When the operating line touches (reaches) the equilibrium line X1´=0.037 (From graph)
r i A g k / 2 O N g k , ' Y
operating Line (Ls/Vs) Min
0.015
0.01
' 0.005 Y2
0 X02'
0.01
0.02
0.03
',
0.04 X1 max
0.05
X' kg NO2/kg gel
29
CONTD. '
Y1
'
X 1
L S
'
'
Y 2
'
V S'
'
X 2 '
L
'
'
Y1
' S min
X
Y 2
' 1
' 2
X
'
V s
V1
'
L S (min) '
L S
0.5 x
B1
1
X
X
0.5
Y 1'
0 1
1
'
xV S
0.0242
0.488 kgAir / s
0.268 kg gel / s
2 x0.268 ' 2
0.0036
0.037
1
0.536 kg gel / s '
' 1
0.0242
'
xV S
V
' S
'
Y1
Y 2 '
L S
0.00
0.488 x
0.0 24
0.0036
0.536
0.0186 kg NO 2 / kg gel
30
0.03
'
Y1
equilibrium curve
0.025
0.02 r i A g k / 2 O N g k , ' Y
operating Line (Ls/Vs) Min
0.015
0.01
0.005
'
Y2
0 0 X2'
0.01
0.02
0.03
X' kg NO2/kg gel
0.04 X1', max
0.05 31
Cocurrent
COCURRENT FLOW
Gel+NO 2
Air/NO2
Air/NO2
Gel
Y1’=0.024
Y2’=0.0036 X1’=0.00
For (LS)min, (X2’)max can be found be drawing the operating line reaching 0.03
the equilibrium line
(X2’)max=0.0034
0.025
equilibrium curve
X1', Y1'
0.02 r i A g k / 2 O N g k , ' Y
operating Line Slope (-Ls/Vs) Min
0.015
0.01
0.005 ',
X2 max,Y2'
0 0
0.01
0.02
0.03
X' kg NO2/kg gel
0.04
0.05
32
SOLUTION, CONTINUED '
Y1
Y 2
'
X '1
X '2
L S
'
V S'
'
'
L
Y1
' S min
L'S
'
Y 2
X 2' ,max
X 1'
2 x 2.957 X 1' 0
V S'
0.0242
0.0036
0.0034
0
x0.488
2.957 kg gel / s
5.92 kg gel / s '
X 2'
'
xVS
Y1
0.488 x
'
Y 2
L'S 0.024
0.0036
5.92
0.00168 kg NO 2 / kg gel
To reach the same degree of removal of NO2, countercurrent flow is much more effective compared to cocurrent The amount of gel needed for cocurrent flow (5.92 kg/s) is about 11 times of that needed if countercurrent flow is used (0.536 kg/s)
33
EQUILIBRIUM-STAGE OPERATION
In many instances mass transfer devices are assembled by interconnecting individual units (stages)
The material passes through each one of these stages Two material streams moves countercurrently (cascades) In each stage the two streams are contacted, mixed, and then separated As the stream moves between stage they come close to equilibrium conditions If the leaving streams from a certain stage is in equilibrium , this stage is an ideal stage If the stage are connected cocurrently they represent a single stage Batch mass-transfer operations are also a single stage 34
STAGE-OPERATIONS V1 VS Y1
L0 LS X0
Stage 1
V2 VS Y2
L1 LS X1
Stage 2
n
Stage N-1
VN VS YN
LN-1 LS XN-1
Stage N
VN+1 VS YN+1
LN LS XN
The flow rate and composition of each stream are numbered corresponding to the effluent from a stage
X2 is the mole ratio in stream leaving stage 2 YN is the mole ratio of stream leaving For ideal stages, the effluents from each stage are in equilibrium Y2 is in equilibrium with X 2 and so on The cascade has the characteristics of the countercurrent process with operating line goes through points (X 0, Y1) and (XN, YN+1)
Thecas
35
NUMBER OF IDEAL STAGES
Number of ideal stages can be determined graphically (for two component systems)
equilibrium curve
Y1 X0, Y 1
Y Y2 Y3 YN+1 XN, Y N+1
36
NUMBER OF STAGES
For linear equilibrium line (Y i=mXi), analytical solution is possible (Kremser):
Define Absorption factor (ratio of the slope of the operating line to the slope of the equilibrium L line): A S
mV S
For transfer from L to V (Stripping) ln N
N
X 0
YN
1
/ m
X N
YN
1
/ m
ln 1 / A X 0 X N
X N Y N
1
1
A
For transfer from V to L (Absorption)
A
ln A
A
1
1
N
N
Y N
1
Y1
mX 0 mX 0 ln A
Y N Y1
1
Y 1 mX 0
1
1
1
A
A
A
1
A
1 37
EXAMPLE:
A flue gas flows at a rate of 10 kmol.s at 298 K and 1 atm with a SO 2 content of 0.15 mole%. Ninety percent of sulfur dioxide is to be removed by absorption with pure water at 298° K. The design water flow rate will be 50% higher than the minimum. Under these conditions, the equilibrium line is: Yi
10 X i
38
SOLUTION
L0, LS, x0, X0
Given Data
T=298° K y1=yN+1=0.0015 x2=x0=0 m=10 V1=10 kmol/s LS=1.5 (LS) min
V1, VS, y1, Y1
1 L2, LS, x2, X2
V2, VS, y2, Y2
Single Stage L1, LS, x1, X1
C a s c a d e
2 3 N-1 N
V1, VS, y1, Y1 LN, LS, xN, XN
VN+1, VS, yN+1, YN+1
39
SOLUTION
y1=yN+1=0.0015Y1=(YN+1)cascade=0. 0015/(1-0.0015)=0.001502 Y2=(Y1)cascade =(1-0.9)*Y1=0.00015 x2=(x0)cascade =0 Vs=V1(1-y1)=10*(1-0.0015)=9.985 kmol/s
From the graph
X1, max=0.00015 (LS /VS)min=(Y1-Y2)/(X1, max-X2) =9 LS min=9*9.985=89.9 kmol/s LS =1.5 Ls min=134.8 kmol/s =134.8*18=2426 kg/s LS /VS=13.5 X1=X2+(Y1-Y2)/(LS /VS) =10-4 mol SO2 /mol H2O = 10-4 *64/18=0.356x10-3 = 0.356 g SO2 /kg H2O
0.002
0.0018
0.0016
Equilibrium
Y1 0.0014
0.0012 Y LS /V
0.001
0.0008
0.0006
(LS /VS)
0.0004
0.0002 Y2
X1
X1 max
0 X02
0. 00 00 2
0 .0 00 04
0 .0 00 06
0 .0 00 08
0 .0 001
0 .0 00 12
0 .0 00 14
0 .0 00 16
0 .0 00 18
X
40
0 .0 00 2
